A person of 60 Kg mass is in a lift which is coming down such that the man exerts a force of 150 N on the floor of the lift. Then the acceleration of the lift is (g=10 `ms^(-2)`)

Options

A. `40.0 ms^(-2)`
B. `15.0 ms^(-2)`
C. `22.5 ms^(-2)`
D. `7.5 ms^(-2)`

Answer

To solve the problem, we need to analyze the forces acting on the person in the lift. Here’s a step-by-step solution:

### Step 1: Identify the forces acting on the person
The forces acting on the person are:
1. The gravitational force (weight) acting downwards, which is given by \( W = mg \).
2. The normal force (N) exerted by the floor of the lift acting upwards.

### Step 2: Calculate the weight of the person
Given:
- Mass of the person, \( m = 60 \, \text{kg} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

The weight of the person is calculated as:
\[ W = mg = 60 \, \text{kg} \times 10 \, \text{m/s}^2 = 600 \, \text{N} \]

### Step 3: Analyze the situation
Since the person exerts a force of 150 N on the floor of the lift, this normal force is less than the weight of the person (600 N). This indicates that the lift is accelerating downwards.

### Step 4: Apply Newton's second law
Using Newton's second law, we can set up the equation for the forces acting on the person:
\[ N = mg - ma \]
Where:
- \( N \) is the normal force (150 N),
- \( mg \) is the weight of the person (600 N),
- \( ma \) is the force due to the acceleration of the lift.

### Step 5: Substitute the known values
Substituting the known values into the equation:
\[ 150 = 600 - 60a \]

### Step 6: Solve for acceleration (a)
Rearranging the equation to solve for \( a \):
\[ 60a = 600 - 150 \]
\[ 60a = 450 \]
\[ a = \frac{450}{60} = 7.5 \, \text{m/s}^2 \]

### Conclusion
The acceleration of the lift is \( 7.5 \, \text{m/s}^2 \) downwards.

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